Level Order Tree Traversal in Python (Time Complexity: O(n)) /homework

Given a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

Example:

Input: [3, 9, 20, null, null, 15, 7]
    3
   / \
  9  20
    /  \
   15   7

Output: 
[
  [3],
  [9, 20],
  [15, 7]
]

Understanding the Problem

The core challenge of this problem is to traverse a binary tree level by level, from left to right. This type of traversal is known as a level order traversal. It is commonly used in scenarios where we need to process nodes on the same level before moving to the next level, such as in breadth-first search (BFS).

Potential pitfalls include not handling null nodes correctly and not maintaining the correct order of nodes at each level.

Approach

To solve this problem, we can use a queue data structure to facilitate the level order traversal. The queue will help us process nodes level by level.

Here is a step-by-step approach:

Initialize an empty list to store the result.
Use a queue to keep track of nodes at each level. Start by adding the root node to the queue.
While the queue is not empty, do the following:
- Initialize an empty list to store the values of nodes at the current level.
- Get the number of nodes at the current level (i.e., the size of the queue).
- For each node at the current level, do the following:
  - Remove the node from the queue and add its value to the current level list.
  - Add the node's left and right children to the queue (if they exist).
- Add the current level list to the result list.

Algorithm

Here is a detailed breakdown of the algorithm:

Initialize an empty list result to store the final level order traversal.
Check if the root is null. If it is, return an empty list.
Initialize a queue and add the root node to it.
While the queue is not empty:
- Initialize an empty list current_level to store the values of nodes at the current level.
- Get the number of nodes at the current level by checking the size of the queue.
- For each node at the current level:
  - Remove the node from the queue and add its value to current_level.
  - If the node has a left child, add it to the queue.
  - If the node has a right child, add it to the queue.
- Add current_level to result.
Return result.

Code Implementation

from collections import deque

# Definition for a binary tree node.
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def levelOrder(root):
    # Result list to store the level order traversal
    result = []
    
    # If the root is None, return an empty list
    if not root:
        return result
    
    # Initialize a queue and add the root node to it
    queue = deque([root])
    
    # While the queue is not empty
    while queue:
        # List to store the current level's nodes
        current_level = []
        # Number of nodes at the current level
        level_length = len(queue)
        
        # Iterate over all nodes at the current level
        for _ in range(level_length):
            # Pop the node from the queue
            node = queue.popleft()
            # Add the node's value to the current level list
            current_level.append(node.val)
            
            # Add the node's children to the queue
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
        
        # Add the current level list to the result list
        result.append(current_level)
    
    # Return the result list
    return result

Complexity Analysis

The time complexity of this approach is O(n), where n is the number of nodes in the binary tree. This is because we visit each node exactly once.

The space complexity is also O(n) due to the space required to store the result and the queue.

Edge Cases

Some potential edge cases include:

An empty tree (root is null). The expected output is an empty list.
A tree with only one node. The expected output is a list with one sublist containing the single node's value.

Testing

To test the solution comprehensively, consider the following test cases:

An empty tree: levelOrder(None) should return [].
A tree with one node: levelOrder(TreeNode(1)) should return [[1]].
A tree with multiple levels: levelOrder(TreeNode(3, TreeNode(9), TreeNode(20, TreeNode(15), TreeNode(7)))) should return [[3], [9, 20], [15, 7]].

Thinking and Problem-Solving Tips

When approaching such problems, it is helpful to:

Understand the type of traversal required (e.g., level order, in-order, pre-order, post-order).
Consider using appropriate data structures (e.g., queue for level order traversal).
Break down the problem into smaller steps and solve each step incrementally.
Practice similar problems to strengthen your understanding and problem-solving skills.

Conclusion

Level order traversal is a fundamental technique in tree traversal, useful in various applications such as breadth-first search. Understanding and implementing this traversal helps in solving more complex tree-related problems.

Practice and explore further to enhance your problem-solving skills and deepen your understanding of tree data structures.